Help! - Calculus

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Over this summer I've been trying to self-educate myself in calculus (or, rather, education with the help of the internet), among other things. I do not quite understand how this site manages to jump from the last step to the second last step.



I am probably embarrassing myself by asking, but I've tried doing the division (albeit at twelve at the morning and I am notoriously prone to making minor mistakes).

Thanks for consideration.

As X=infinity, then the 1/x2 and 5/x2 = 0. Cross those out and you have -1 divided by 8. (Funny part is I don't know calculus but its intuitively obvious)

Any finite number over infinity equals zero (approximately). Just imagine 1/x^2 as being 1/(100^2), then 1/(1,000^2), and so on, and you can clearly see that as x gets large (as it approaches infinity) 1/x^2 becomes small enough to be negligible.

Yes, a good question. The concept of limits and infinity is essential to calculus, yet usually not explained or proven until later in mathematical studies. For now, you're probably better off just using the advice given in the comments above. Both explain what is going on quite well.

The bigger x gets (and the fact that x is squared merely accelerates the rate at which x increases), the smaller the value of the fraction (1/1, 1/4, 1/9, 1/16... is the sequence that results as you work your way up the integers; substitute 5 for 1 in the case of the other fraction). Very quickly you have numbers like 1 over 16.8 million, and then 1 over infinity, which is to say (for reasons explained in higher maths) zero.

Once you get used to this idea, solving limits becomes straight-forward.

The definition of a limit is the value of a function as x approaches a certain number. Depending on the type of function, the actual y-value of the function at x can be different from the limit, or the limit may exist when the y-value does not.

In this case, the notation lim (x->infinity) indicates that the graph of f(x) approaches a horizontal asymptote at the line y=L (L being the value of the limit). This means that the greater the value for x, the closer the graph of f(x) gets to the line y=L. Of course, the value of f(x) never reaches L, it just gets infinitely close to it.

Now, about your problem. The website has already simplified the function for you. Now all that remains is to substitute the number n (limit as x->n); however, we can't exactly do that since n = infinity, and is not a discrete number. However, we can assume that x is an ever-increasing number. When you look at it from that perspective:

L = ((1/x^2) - 1)/(8+(5/x^2)) = (0-1)/(8-0)

Basically, since x is an ever increasing number, then 1/x^2 must be decreasing (think: you're taking the number 1 and dividing it into smaller and smaller fractions.). Therefore, since 1/x^2 decreases as x increases, then at infinity 1/x^2 = 0. The same goes for 5/x^2.

Now that weve done that, the rest is easy:

L = (0-1)/(8-0) = -1/8

And there you have it. Not meaning to brag, but this stuff is pretty easy for me. I took AP Calculus I in my senior year of high school and ended up getting a 4 out of 5 on the AP Exam. This basically means I can skip Calc I in college, and take Calc II my freshman year. So if you have any more questions let me know.

I took a little time, figuring out how to line up the "equals" signs in the LaTeX stuff:



Note that there is a slight leap, in the expansion, where the assumption is made that one can distribute a limit across the top and bottom halves of the division. In this instance, this is valid, as both the denominator and the numerator are strictly convergent to finite values, and the numerator is not zero. It is NOT a valid step, should either still be tending to infinity, OR if both happen to converge to zero.

If you download the little Java laeqed utility (from http://www.thrysoee.dk/laeqed/), and save the image above, you will find you can edit it.

Thanks, everyone, for your patience in helping a semi-numerate.